smc

Let $f(x)=x^2+px+q$. Then, $x^{2}f(\frac{1}{x})$ equals $q{x}^2+px+1$ and has roots which are the reciprocals of those of $f(x)$. Thus, by Vieta's Formulas, the sum of the roots of $x^{2}f(\frac{1}{x})$ (and thus the sum of the reciprocated roots of $f(x)$) is $\boxed{-\frac{p}{q}}$.